Localization

Recall that given an integral domain, there is a canonical way to construct the “smallest field containing it”, its field of fractions. Here, we will generalize this construction to arbitrary rings.

We let A be a fixed ring throughout.

Definition.

A subset is said to be multiplicative if

;

if , then .

Our objective is to construct a ring by inverting all elements of S. We will, in fact, define this via its universal property.

Definition.

For a multiplicative , its localization comprises of a pair

where is a ring and is a ring homomorphism such that is a unit for any , with the following property.

For any pair , where is a ring and is a homomorphism such that is a unit for any , there is a unique ring homomorphism such that .

In summary, any ring homomorphsim which maps S into the units of B must factor through the localization.

By definition, we obtain a bijection for any ring B

Intuitively, we may imagine as the “smallest” ring extension of A in which every element of S becomes a unit. [Technically, this is wrong since is not injective in general, but this is just for intuition.]

Exercise A

Prove that if and are both localizations, there is a unique ring isomorphism such that .

Prove that if A is a domain and , then is the field of fractions of A.

Concrete Definition

We will prove that exists by construction.

Proposition 1.

Take with the equivalence relation:

if there exists such that .

and be the set of its equivalence classes; write for the equivalence class containing . Then is a ring under the following operations.

.

Proof

The proof is tedious but straight-forward. For example, for transitivity, if and , then there exist such that

To prove that addition and product are well-defined, suppose so that for some ; we wish to show and . Now

and we are done. It remains to show that these operations turn into a ring, which we will leave to the reader. ♦

Proposition 2.

The ring together with the homomorphism

gives the localization.

Proof

Note that is a unit for each since . Now for any ring B and homomorphism such that is a unit for each , we set

This map is well-defined since if then for some so and since are all units we have .

It is also straight-forward to show that f is a ring homomorphism and .

To show that f is unique, we have for all . Hence for all , we have

♦

Note

From this concrete description of , we see that

if S has no zero-divisors of A, then is injective;

in particular if A is an integral domain and , then , the field of fractions of A;

as a consequence, if A is an integral domain so is .

Some Properties of Localizations

Lemma 1.

If and are multiplicative subsets and is a ring homomorphism such that , we obtain a ring homomorphism

.

Proof

Easy exercise: can be done directly or by universal property. ♦

Some common instances of localizations are as follows.

Example 1: Inverting One Element

The easiest way to get a multiplicative set is to pick and set . We write for the resulting .

For example if , then

,

the -subalgebra of generated by .

Exercise B

Prove that , where is the polynomial ring.

Example 2: Localization at a Prime

Let be a prime ideal. Then is multiplicative by the definition of prime ideals. We write , the localization of A at the prime ideal .

For example, suppose . Then

Usually denotes the ring of 2-adic integers (which we will cover much later). Hence we will write for the first example and for the second (note the brackets).

Ideals of Localized Ring

General Case

Given a general ring homomorphism , we can “push” an ideal through f to obtain an ideal of . Recall that the product notation means we take the set of all finite sums

One can also think of it as the ideal of B generated by .

Conversely, an ideal of B gives us an ideal of A. For convenience we write for and for if f is implicit.

Note that this definition of is consistent with our earlier definition of where M is an A-module.

Main Case of Interest

Now we apply this to our map , .

Lemma 2.

is exactly the set of elements of which can be written as for some .

Note

It is possible for to hold in with but .

Proof

Let be the set of elements as described. It suffices to show:

,

is an ideal of .

For the first claim, let . Then so . Also . Hence .

For the second, let , and . Then

since and . So is an ideal of . ♦

The following gives a relationship between ideals of and those of .

Proposition 3.

For an ideal , let , an ideal of A. Then

.

Proof

(⊆) Suppose , i.e. . Since is an ideal of we have .

(⊇) Pick where . Then so . We see that . ♦

Note

Hence we obtain the following correspondence.

One can imagine this as: the ideals of form a subset of those of A. Clearly this subset is proper in general (e.g. take and ).